5: Colligative Properties

Correct answer: A. Keep the solution sufficiently dilute

Explanation: The osmotic-pressure method is based on the dilute-solution relation π = CRT, or more generally π = iCRT. The solution should therefore be sufficiently dilute so that it behaves approximately ideally and the measured pressure can be related reliably to concentration. Temperature must also be controlled, but making the solution dilute is the key precaution listed here. Colour and an open vessel are irrelevant or undesirable.

Correct answer: D. Al₂(SO₄)₃

Explanation: The van’t Hoff factor represents the effective number of particles formed from one formula unit of solute. NaCl gives two ions, Na⁺ and Cl⁻, so i = 2. Glucose does not ionise, so i = 1. BaCl₂ gives three ions, Ba²⁺ and two Cl⁻, so i = 3. Al₂(SO₄)₃ gives two Al³⁺ ions and three SO₄²⁻ ions, a total of five particles; therefore it has the highest ideal van’t Hoff factor, i = 5.

Correct answer: A. The relative lowering of vapour pressure is related to the mole fraction of the solute.

Explanation: For a solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p° − p)/p° = xsolute, where p° is the vapour pressure of the pure solvent, p is the vapour pressure of the solution, and xsolute is the solute mole fraction. Thus the effect depends on the number fraction of solute particles, not on colour. The solute need not be volatile, and the temperature need not be zero.

Correct answer: A. Association of solute particles

Explanation: The van’t Hoff factor compares the actual effective number of solute particles with the number expected ideally. When solute molecules associate, such as by dimerisation, two or more original particles combine into fewer particles. Consequently, the observed freezing-point depression is smaller than the ideal value and i becomes less than one. Dissociation or ionisation produces extra particles and usually gives i greater than one, so option A is the only suitable explanation.

Correct answer: A. Because its dissociation is partial

Explanation: A weak electrolyte establishes an equilibrium between undissociated molecules and ions in solution; it does not convert completely into ions. Therefore, the number of particles actually present is smaller than the number predicted for complete dissociation. Since colligative properties depend on the effective particle count, the observed van’t Hoff factor is below the ideal value. The extent of dissociation can also vary with concentration, temperature, and the nature of the solvent.

Correct answer: A. Water moves out of the cell.

Explanation: A hypertonic solution has a higher concentration of osmotically active solute particles, and therefore a lower water potential, than the fluid inside the blood cell. Across the selectively permeable cell membrane, water moves outward toward the side with the greater effective solute concentration. The cell loses water and shrinks, a process called crenation in red blood cells. This is why isotonic saline is used for safe intravenous treatment.

Correct answer: A. Water enters the cell.

Explanation: A hypotonic solution contains fewer effective solute particles and has a higher water potential than the fluid inside the blood cell. Because the cell membrane allows water to pass more readily than many solutes, water moves into the cell by osmosis. The cell swells and, if water entry is excessive, may burst; bursting of a red blood cell is called haemolysis. The result is opposite to what occurs in a hypertonic solution.

Correct answer: A. The molar mass will be incorrect and generally lower

Explanation: Dissociation increases the number of particles in solution, so the observed colligative effect becomes larger than expected for the original formula units. If this enhanced effect is treated as though no dissociation occurred, the calculated number of moles is overestimated. Since molar mass equals mass divided by moles, the calculated molar mass is generally lower than the true value. Thus, option A is correct.

Correct answer: C. 1.5

Explanation: For abnormal molar mass, the observed molar mass is related to the true molar mass by Mobserved = Mtrue/i. Substituting the given values gives 80 = 120/i. Rearranging, i = 120/80 = 1.5. A value greater than one indicates that the solute has produced more particles, usually because of dissociation. Therefore, option C is correct.

Correct answer: C. 100 g mol⁻¹

Explanation: For a non-dissociated solute, boiling-point elevation is ΔTb = Kb m. Therefore, m = 0.156/0.52 = 0.30 mol kg⁻¹. The solvent mass is 100 g = 0.100 kg, so moles of solute are 0.30 × 0.100 = 0.030 mol. The molar mass is mass/moles = 3/0.030 = 100 g mol⁻¹. Hence option C is correct.

Correct answer: B. 0.10 M

Explanation: Isotonic solutions have equal osmotic pressure, and osmotic pressure is π = iCRT. Glucose is a non-electrolyte, so i = 1 and its effective particle concentration is 1 × 0.20 = 0.20 M. Completely dissociated NaCl gives two ions, so i = 2. Therefore, 2C = 0.20, giving C = 0.10 M. Hence option B is correct.

Correct answer: C. Greater than the osmotic pressure

Explanation: In normal osmosis, solvent passes through a semipermeable membrane from the dilute solution toward the concentrated solution. In reverse osmosis, pressure is applied on the concentrated seawater side to force solvent in the opposite direction. The applied pressure must exceed the osmotic pressure; pressure equal to osmotic pressure only stops net osmosis. Therefore, option C is correct.

Correct answer: C. 2.5

Explanation: For a solution, the freezing-point depression is ΔTf = iKf m. For the same solvent and molality, the non-dissociated value corresponds to i = 1. Thus, the ratio of the actual depression to the non-dissociated depression is exactly i. Since the actual value is 2.5 times the expected value, i = 2.5. Therefore, option C is correct.

Correct answer: A. It can be measured even for very dilute solutions

Explanation: Large biomolecules have very high molar masses, so a given mass produces very few moles and usually requires a very dilute solution. In such solutions, boiling-point elevation or freezing-point depression may be too small to measure accurately. Osmotic pressure, however, can be measured at ordinary or low concentrations and is proportional to molar concentration through π = CRT. Therefore, A is correct.

Correct answer: C. It becomes double

Explanation: The freezing-point depression is given by ΔTf = iKf m. If the effective number of solute particles doubles while the solvent, Kf, and stated molality basis remain unchanged, the van’t Hoff factor i doubles. Since ΔTf is directly proportional to i, the freezing-point depression also doubles. Thus, option C is the correct consequence of the colligative-property relation.

Correct answer: A. 0.025 M

Explanation: For a non-dissociated solute, the van’t Hoff factor is i = 1, so the osmotic-pressure equation is π = CRT. First convert the temperature: 25 °C = 298 K. Therefore, C = π/(RT) = 0.615/(0.082 × 298) = approximately 0.0252 mol L⁻¹. Hence the nearest option is 0.025 M, option A.

Correct answer: B. 2

Explanation: For boiling-point elevation, use ΔTᵦ = iKᵦm, where i is the van’t Hoff factor, Kᵦ is the ebullioscopic constant, and m is molality. Substitution gives 0.104 = i × 0.52 × 0.1. Since 0.52 × 0.1 = 0.052, i = 0.104/0.052 = 2. Therefore, the solution produces twice the ideal number of particles and option B is correct.

Correct answer: A. Osmotic pressure, because it can be measured at room temperature even for very dilute solutions

Explanation: For a high-molar-mass protein, a given mass produces very few moles of solute. Consequently, boiling-point elevation, freezing-point depression, and vapour-pressure lowering are extremely small and difficult to measure accurately. Osmotic pressure is appreciable even in dilute solutions and can be measured at room temperature, avoiding protein decomposition. Therefore, A is correct.

Correct answer: B. 0.10 mol

Explanation: For a dilute solution containing a non-volatile solute, relative lowering of vapour pressure is approximately equal to the solute mole fraction: Δp/p⁰ ≈ x₂. Let n₂ be the solute moles. Then 0.10 = n₂/(0.90 + n₂). Solving gives 0.10(0.90 + n₂) = n₂, so n₂ = 0.10 mol. Therefore, B is correct.

Correct answer: D. Four times

Explanation: Freezing-point depression is given by ΔTf = iKf m. A non-dissociated solute has i = 1. Complete dissociation of AlCl₃ produces one Al³⁺ ion and three Cl⁻ ions, giving four particles and i = 4. Since molality and Kf are the same in both cases, the AlCl₃ solution has four times the freezing-point depression. Option D is correct.

Correct answer: C. Glucose and AlCl₃

Explanation: Osmotic pressure is π = iCRT. At the same concentration and temperature, its value depends on the van’t Hoff factor i. Glucose and urea are non-electrolytes with i = 1; NaCl and KCl each give i = 2; CaCl₂ and MgCl₂ each give i = 3; AlCl₃ gives i = 4. Therefore, glucose versus AlCl₃ has the largest difference, 4 − 1 = 3, so C is correct.

Correct answer: B. Dimer-like association

Explanation: For a dilute solution, osmotic pressure is π = iCRT. At fixed concentration and temperature, the observed pressure being half the non-dissociated value means i = 0.5 instead of 1. A van’t Hoff factor below one shows that solute particles have associated, reducing the number of independent particles; dimer formation is a typical example.

Correct answer: D. AlCl₃

Explanation: For a dilute solution, osmotic pressure is π = iCRT. Since C and T are identical for all solutions, the solution with the largest van’t Hoff factor i has the greatest osmotic pressure. Urea gives i=1, KBr gives i=2, BaCl₂ gives i=3, and complete dissociation of AlCl₃ gives Al³⁺ plus three Cl⁻ ions, so i=4 and option D is correct.

Correct answer: B. 0.2 m

Explanation: The boiling-point elevation equation is ΔTb = iKb m. Rearranging gives m = ΔTb/(iKb). Substituting the given values, m = 0.208/(2 × 0.52) = 0.208/1.04 = 0.20 mol kg⁻¹. The van’t Hoff factor must be included because the solute produces two effective particles per formula unit.

Correct answer: A. Increase the concentration of solute particles

Explanation: For a dilute solution, osmotic pressure is represented by π = iCRT, where i is the van’t Hoff factor, C is the concentration of solute particles, R is the gas constant, and T is absolute temperature. When temperature is fixed and the nature of the solute is unchanged, increasing C increases π directly. Colour, naming, and removal of the membrane do not increase the osmotic pressure of the original solution.